Surds, Indices and Simplification — SPI Mathematics Formulas and Traps

Laws of Indices

Multiplying, dividing and raising powers

Laws of indices

a^m \times a^n = a^{m+n} \qquad a^m \div a^n = a^{m-n} \qquad (a^m)^n = a^{mn} \qquad (ab)^n = a^n b^n

Zero and negative indices

a^0 = 1 \ (a \ne 0) \qquad a^{-n} = \frac{1}{a^n} \qquad \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^{n}

Common traps

Adding indices for a power of a power

(a^2)^3

is

a^6

, not

a^5

. Add the indices only when two powers of the same base are multiplied.

Joining different bases

2^3 \times 3^2 = 8 \times 9 = 72

, not

6^5

. The laws that add or subtract indices work only for one base.

Spreading a power over a sum

(a + b)^2

is not

a^2 + b^2

:

(2 + 3)^2 = 25

, but

4 + 9 = 13

. A power spreads over a product, never over a sum.

Reading a zero index as 0

a^0

is 1, not 0, for every non-zero a. Only

0^0

is left undefined.

Reading a negative index as a negative number

2^{-3} = \frac{1}{8}

, not

-8

. The minus sign in the index turns the number into its reciprocal.

Losing the sign of a negative base

\left(-\frac{3}{4}\right)^2 = \frac{9}{16}

is positive, but

\left(-\frac{3}{4}\right)^3 = -\frac{27}{64}

is negative. Check whether the power is even or odd.

Fractional Indices and Exponential Equations

Learn this subtopic in the notes

Fractional indices as roots

Fractional index

a^{\frac{1}{n}} = \sqrt[n]{a} \qquad a^{\frac{p}{q}} = \left(\sqrt[q]{a}\right)^{p}

Solving equations in powers

Equal powers of one base

a^x = a^y \Rightarrow x = y \quad (a > 0,\ a \ne 1)

Common traps

Swapping the root and the power

In

8^{\frac{2}{3}}

the 3 is the root and the 2 the power:

(\sqrt[3]{8})^2 = 4

. Reading it as

8^{\frac{3}{2}}

gives about 22.6.

Forgetting to turn a fraction over

\left(\frac{9}{16}\right)^{-\frac{1}{2}}

is

\frac{4}{3}

, not

\frac{3}{4}

. A negative index flips the fraction before the root is taken.

Miscounting the decimal places

0.0016 = (0.2)^4

because

2^4 = 16

and four places of 0.1 give four decimal places. Count places before writing a decimal as a power.

Matching indices on different bases

From

4^x = 8

you cannot say

x = 2

. Make the bases equal first,

2^{2x} = 2^3

, and only then match the indices.

Treating a difference of powers as one power

3^{x+2} - 3^x

is not

3^2

. Take out

3^x

: it equals

3^x(9 - 1) = 8 \times 3^x

.

Missing the answer x = 1

Matching indices needs a base that is not 1. In

x^{2x} = x^6

the indices give

x = 3

, but

x = 1

also works, because 1 to any power is 1. Check which values the question allows.

Surds and Rationalising

Learn this subtopic in the notes

Simplifying and comparing roots

Roots as powers

\sqrt{ab} = \sqrt{a}\,\sqrt{b} \qquad \sqrt[n]{a} = a^{\frac{1}{n}}

Rationalising the denominator

Conjugate pair

(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b \qquad (\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b

Square root of a surd and endless roots

Square root of a surd and an endless root

\sqrt{(m + n) + 2\sqrt{mn}} = \sqrt{m} + \sqrt{n} \qquad x = \sqrt{k + x} \Rightarrow x^2 - x - k = 0

Common traps

Splitting a root over a sum

\sqrt{9 + 16} = \sqrt{25} = 5

, not

3 + 4 = 7

. A root splits over a product, never over a sum.

Getting the decimal places wrong

\sqrt{0.0049} = 0.07

, not 0.7. The number under the root has four decimal places, so its root has two.

Adding unlike surds

\sqrt{2} + \sqrt{3}

is not

\sqrt{5}

: it is about

1.41 + 1.73 = 3.14

, while

\sqrt{5}

is about 2.24.

Multiplying only the denominator

Multiplying the bottom alone by the conjugate changes the value. Multiply the top by the same factor.

Losing a negative denominator

For

2 - \sqrt{7}

the product with its conjugate is

4 - 7 = -3

. Carry that minus sign into every term of the answer.

Using the wrong square identity

x^2 + y^2 = (x + y)^2 - 2xy

, but

x^2 + xy + y^2 = (x + y)^2 - xy

. Expand

(x + y)^2

and take away only what is extra.

Not making the root term 2 root b first

In

11 + \sqrt{120}

, write

\sqrt{120} = 2\sqrt{30}

before looking for m and n. Then

m + n = 11

,

mn = 30

, giving

\sqrt{6} + \sqrt{5}

.

Keeping the negative root

The quadratic for an endless root has one negative solution. A square root cannot be negative, so that solution is rejected.

Putting the smaller number first

\sqrt{4 - 2\sqrt{3}}

is

\sqrt{3} - 1

, not

1 - \sqrt{3}

, which is negative.

Surds, Indices and Simplification formulas

Laws of Indices

Multiplying, dividing and raising powers

Common traps

Fractional Indices and Exponential Equations

Fractional indices as roots

Solving equations in powers

Common traps

Surds and Rationalising

Simplifying and comparing roots

Rationalising the denominator

Square root of a surd and endless roots

Common traps

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